If a $13 k g$ $13 kg$ object moving at $400 \frac{m}{s}$ $400 m/s$ slows down to a halt after moving $20 m$ $20 m$ , what is the friction coefficient of the surface that the object was moving over?

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Answer 1 · 2018-03-31T12:42:35

The coefficient of friction is $= 408.2$ $=408.2$

Apply the equation of motion to find the acceleration

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

The initial velocity is $u = 400 m s^{- 1}$ $u=400ms^-1$

The final velocity is $v = 0 m s^{- 1}$ $v=0ms^-1$

The distance is $s = 20 m$ $s=20m$

The acceleration is

$a = \frac{v^{2} - u^{2}}{2 s} = \frac{0 - 400^{2}}{2 \cdot 20} = - 4000 m s^{- 2}$ $a=(v^2-u^2)/(2s)=(0-400^2)/(2*20)=-4000ms^-2$

The mass of the object is $m = 13 k g$ $m=13kg$

According to Newton's Second Law of Motion

$F = m a$ $F=ma$

The force of friction is $F_{r} = 13 \cdot 4000 = 52000 N$ $F_r=13*4000=52000N$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The normal reaction is $N = m g = 13 \cdot 9.8 = 127.4 N$ $N=mg=13*9.8=127.4N$

The coefficient of kinetic friction is

$μ_{k} = \frac{F_{r}}{N} = \frac{52000}{127.4} = 408.2$ $mu_k=F_r/N=52000/127.4=408.2$

If a 13kg13 kg object moving at 400ms400 m/s slows down to a halt after moving 20m20 m, what is the friction coefficient of the surface that the object was moving over?